difference between two control statements for an if

[cooper1200]

i have this code

#include <stdio.h>
#include <stdlib.h>

#define MAX_NUMBER 4000000

void fill_fib_nums(int fib_nums[], int num_elements);
int even_sum(int fib_nums[], int num_elements);

int main()
{
    int answer, num_elements, fib_nums[33];

    fib_nums[0] = 1;
    fib_nums[1] = 1;
    num_elements = sizeof(fib_nums) / sizeof(fib_nums[0]);
    fill_fib_nums(fib_nums, num_elements);
    answer = even_sum(fib_nums, num_elements);
    printf("the some of the even numbers in the fibonachi sequence that are less than 4,000,000 is %d\n", answer);

    return 0;
}

void fill_fib_nums(int fib_nums[], int num_elements)
{
    int i, j, next_num = 2;

    for (i = 0, j = 1; next_num < num_elements; i++, j++, next_num++)
    {
        fib_nums[next_num] = fib_nums[i] + fib_nums[j];//add the previous two numbers in the array together and store the result in fib_nums[next_num]
    }

}

int even_sum(int fib_nums[], int num_elements)
{
    int i, answer;

    for (i = 0; i < num_elements; i++)
    {
        if (fib_nums[i] % 2 == 0) // run if fib_nums[i] is even
        {
            answer += fib_nums[i];
        }
    }

    return answer;
}

the above doesn't work. however if i change the if statement on line 40 to

 if (fib_nums[i] % 2 == 0)

it does.

i thought that

if(!some statement that equates to 0)

is the same as

if (some statement that equates to 0 == 0)

coop

[laserlight]

Uh. Line 40 and what you wrote as its replacement look the same to me.

[cooper1200]

oh bother thats what happends when you cut and paste code after finding the soloution. line 40 was

 if (!fib_nums[i] % 2)

which didn't work

[laserlight]

Oh, I see. This:

if (!fib_nums[i] % 2)

Is equivalent to:

if ((fib_nums[i] == 0) % 2)

[cooper1200]

huh surely

 if ((fib_nums[i] == 0) % 2)

is an error as you can't divide 0

[cooper1200]

i see what your saying i should of written

 if (!(fib_nums[i] % 2))

[GReaper]

huh surely

 if ((fib_nums[i] == 0) % 2)

is an error as you can't divide 0

You can't divide by zero. Dividing zero by anything (other than zero) is certainly not an error.

[flp1969]

As laserlight told you, this expression: !fib_nums[i] % 2 is evaluated as (!fib_nums[i]) % 2. It is important to know the operators precedence and associativity when building an expression: C Operator Precedence.

Precedence is what operator is evaluated first... Associativity is when you have two similar operators, which of them is evaluated first. For exemple:

x = a * b + c * d;

In this expression the operator * has precedence over the operator +, and the later has precedence over the operador =. So this expression can be undestood as:

 x = ( ( a * b ) + ( c * d ) );

Where the multiplications are done before the add and then the assignment is done. Notice you have two multiplications, which one is done first? * operator has associativity from left to right, so (a * b) is done first, then (c * d)...

In the expression !fib_nums[i] % 2 we got 3 operators: !, [] and %. [] has precedence over !, and % has the lesser precedence (it will be evaluated last). So the expression is ( ! ( fib_numbs[i] ) ) % 2.

The expression fib_nums[i] % 2 == 0, == has the lesser precedence, so the expression is evaluated as ( ( fib_nums[i] ) % 2 ) == 0.

[laserlight]

Precedence is what operator is evaluated first...

That's a common misconception. Precedence determines grouping, i.e., how the syntax is to be interpreted. It doesn't determine order of evaluation, except where the syntax implies a particular order of evaluation.

Notice you have two multiplications, which one is done first? * operator has associativity from left to right, so (a * b) is done first, then (c * d)...

That isn't true though. Associativity can determine order of evaluation when it is implied by the grouping imposed by association, but in this case it doesn't: the two subexpressions are separate, unlike say, a * b * c. Therefore, the order of evaluation is not specified, so either multiplication could occur first.

[cooper1200]

so the ! operator has precedence over the % operator?

[laserlight]

Yes.